a) If \(f(x) = \frac{2x-8}{x^2 -2x - 3} \qquad\text{ and }\qquad g(x) = \frac{3x+9}{2x-4}\) find the sum of the values of x where the vertical asymptotes of f(g(x)) are located.
b) What is the horizontal asymptote as x approaches negative infinity of f(g(x))?
+128474 (1) f(g (x) ) means : put g into f
Because the math is messy...I'm going to use WolframAlpha to simplify this ...so f(g(x)) =
4(x - 5) (x - 2) 4 (x- 5)(x - 2) 4x^2 - 28x + 40
____________ = ______________ = ______________ (1)
3(x^2 - 6x - 7) 3 (x - 7) ( x + 1) 3x^2 - 18x - 21
The vertical asymptotes are the x values that make the denominator = 0 these are x = 7 and x = -1
So....their sum = 7 + -1 = 6
(2) As x approaches pos/neg infinity.....y approaches the ratio of the coefficients on the x^2 terms in (1).....that is
y = 4/3 ......and this is the horizontal asymptote in both directions
See the graph here : https://www.desmos.com/calculator/xg6a4ctnqz
